a pitched ball is hit by a batter at a 45 degree angle and just clears the outfield fence, 98m away.?

Question

Assume the fence is the same hieght as the pitch and find the velocity of the ball when it left the bat. neglect air resistance.

i got 31.6 approximately but my teacher says im wrong and the books ays im right am i??

Answer 1

X(t)=v0*cos(45)*t
and
y(t)=v0*sin(45)*t-.5*g*t^2
when y(t)=0 the ball leaves the bat and clears the fence
when x(t)=98, the ball clears the fence

98=v0*cos(45)*t
0=v0*sin(45)*t-.5*g*t^2

for 45 degrees, sin=cos, so
0=98-.5*g*t^2
t=4.47 seconds

v0=31.01 m/s

I used g=9.81 m/s^2

j